VCE Chemistry 2023

Every question from the 2023 VCE Chemistry examination, the last under the previous (2017 to 2023) study design. A few topics, such as flashpoint and the old Data Book tables, are no longer in the course, but most of the paper is still excellent practice. Multiple-choice answers come with a short reason; Section B solutions sit behind a Show worked solution toggle.

How to use this page

• Questions are from the 2023 VCE Chemistry examination, copyright Victorian Curriculum and Assessment Authority (VCAA). Each is summarised briefly here; open the official examination PDF for the full wording, diagrams, spectra and the Data Book.
• Answers are original ExamExplained working. Every multiple-choice answer matches the key in the 2023 Chemistry examination report (Word document), and every Section B calculation was redone and compared with the report. Both files are listed on the VCAA Chemistry examinations page.
• Percentages after multiple-choice answers are the share of students who chose the correct option. In Section B, "From the report" notes summarise what the examiners said students got wrong.

Structure and timing

The paper is 120 marks in 150 minutes (plus 15 minutes reading time), with a scientific calculator and the Data Book. That is 1.25 minutes per mark.

• Section A (30 marks): 30 multiple-choice questions. Aim for about 35 minutes.
• Section B (90 marks): 9 questions (11, 12, 12, 7, 7, 7, 10, 16 and 8 marks). Show working, give states in equations and use sensible significant figures.

Section A: Multiple choice

Q1

The reaction of glucose in human cells. Answer: B - respiration, which is exothermic. (67%)

Q2

Which statement about fuel cells and galvanic cells is correct? Answer: A - both produce heat as well as electrical energy. Galvanic cells can have inert electrodes and porous electrodes too, and electrons never flow through the electrolyte. (40%; 42% chose C.)

Q3

Order of melting points for three triglycerides with the same number of carbons: X has one C=C double bond, Y none and Z three. Answer: A - more C=C bonds mean more kinks, looser packing, weaker attraction and a lower melting point, so Y is highest, then X, then Z. (70%)

Q4

A cell is a secondary cell if? Answer: D - it can operate as both a galvanic cell (discharge) and an electrolytic cell (recharge). (79%)

Q5

Three statements about coenzymes. Answer: D - all three are true: they assist enzymes, bind loosely at the active site and carry electrons or groups of atoms. (42%)

Q6

Which statements apply to HPLC? Answer: C - it is qualitative (retention times) and quantitative (peak areas). Triplets belong to ¹H NMR. (43%)

Q7

Statements about fossil fuels and biofuels. Answer: B - only II: both produce greenhouse gases when burned. Biofuel production can damage the environment, and fossil fuels are not renewable. (73%)

Q8

$K = 4.10$ M⁻² for N₂ + 3H₂ ⇌ 2NH₃. Find $K$ for 2N₂ + 6H₂ ⇌ 4NH₃. Answer: D - doubling the coefficients squares $K$: $4.10^2 \approx 16.8$ M⁻⁴. (52%)

Q9

Coal seam gas and petroleum gas. Answer: D - both are mixtures, both produce CO₂ on combustion and both are fossil fuels. (54%)

Q10

At constant temperature, what always corresponds to a higher yield? Answer: A - an increase in the value of the equilibrium expression (more products relative to reactants). Adding reactant or raising pressure does not always change the yield (for example H₂ + Cl₂ ⇌ 2HCl), and a lower activation energy changes only the rate. (26%; 47% chose B.)

Q11

Units of the initial rate from a product concentration-time graph. Answer: A - M s⁻¹, that is mol L⁻¹ s⁻¹. (63%)

Q12

What does a catalyst change? Answer: A - only the activation energy, not $\Delta H$ or the enthalpy of the reactants. (88%)

Q13

Which molecule has cis and trans isomers? Answer: C - CH₃CH=CHCl: each carbon of the C=C carries two different groups. (59%)

Q14

Effect of raising the temperature on the rates for N₂O₄ ⇌ 2NO₂ ($\Delta H > 0$). Answer: C - both the forward and reverse rates increase (the forward rate more, since it is endothermic). (44%)

Q15

Petrol-powered and hydrogen fuel cell drones. Answer: A - both produce greenhouse gases: CO₂ and H₂O from petrol, and H₂O (a greenhouse gas) from the fuel cell. Both reactions are exothermic, but their energy transformations differ. (41%)

Q16

Number of ¹³C NMR peaks for the ester CH₃CH₂CH(CH₃)COOCH(CH₃)₂. Answer: C - 7: the two CH₃ groups on the O-CH carbon are equivalent, and the other six carbons are all different. (43%)

Q17

Lignite releases 19.0 MJ per tonne and the station is 39% efficient. Mass of lignite for 42.0 MJ of usable energy? Answer: D - total energy needed $= \tfrac{42.0}{0.39} = 107.7$ MJ, so $\tfrac{107.7}{19.0} \approx 5.67$ tonnes. (57%)

Q18

A 1.00 mL sample diluted to 1.00 L gives a peak area of 0.35 (18 ng mL⁻¹ on the calibration curve). Original concentration? Answer: C - multiply by the dilution factor of 1000: 18 000 ng mL⁻¹ $= 18$ µg mL⁻¹. (37%; 45% chose B, forgetting the dilution.)

Q19

Omega-6 fatty acids compared with saturated fatty acids with the same number of carbons. Answer: A - the C=C creates a kink, so the chains pack less closely and the density is lower. (59%)

Q20

Which is always correct in an accurate acid-base titration? All students were awarded the mark: VCAA found no correct or best answer.

Q21

Denaturation. Answer: D - it alters the secondary (and tertiary) structure of a protein, but not the primary structure. (65%)

Q22

Methane and methanol fuel cells. Answer: B - each mole of either fuel gives 1 mol of CO₂ and 2 mol of H₂O (the same greenhouse gases), but methane releases 8 electrons per mole and methanol 6. (41%)

Q23

Semi-structural formula of 2-methylpent-1-ene. Answer: C - CH₂C(CH₃)CH₂CH₂CH₃. (89%)

Q24

Acidic propan-1-ol fuel cell. Answer: A - H⁺ ions produced at the anode flow through the membrane to the cathode. (61%)

Q25

The lead-acid accumulator as a galvanic cell and as an electrolytic cell. Answer: C - on discharge, H⁺ reacts at the cathode (PbO₂ reduction); on recharge, HSO₄⁻ is produced at both electrodes. (44%)

Q26

Number of independent variables in the biodiesel viscosity study. Answer: B - three: the oil, the catalyst and the alcohol. (50%)

Q27

Most valid conclusion from the flow-rate tables. Answer: B - in every case the methanol biodiesel flows faster than the ethanol biodiesel, so it has the lower viscosity. The oil and catalyst comparisons are not consistent across both tables. (45%)

Q28

A cell with S₂O₃²⁻/SO₃²⁻/OH⁻ (Pt electrode) and Mn³⁺/Mn²⁺. Answer: B - the strongest reductant, SO₃²⁻ ($-0.94$ V couple), reacts with Mn³⁺ ($+1.56$ V): $1.56 + 0.94 = 2.50$ V. The Mn³⁺/Mn²⁺ half-cell needs an inert Pt electrode, since Mn metal would react directly with Mn³⁺. (45%)

Q29

Which statement about mass spectrometry is always correct? Answer: D - peak heights are measured relative to the base peak (100%). The molecular mass comes from the molecular ion, not the base peak. (46%)

Q30

1 M H₂O₂, Fe(NO₃)₃ and Fe(NO₃)₂ are mixed. Answer: C - H₂O₂ is both the strongest oxidant (+1.77 V) and the strongest reductant present (the O₂/H₂O₂ couple at +0.68 V lies below Fe³⁺/Fe²⁺ at +0.77 V), so it reacts immediately and its concentration falls. (48%)

Section B

Question 1 (11 marks)

Coconut oil's fatty acids are mainly saturated: caprylic C₈H₁₆O₂ 8%, capric C₁₀H₂₀O₂ 7%, lauric C₁₂H₂₄O₂ 49%, myristic C₁₄H₂₈O₂ 18%, palmitic C₁₆H₃₂O₂ 8%, stearic C₁₈H₃₆O₂ 2%, oleic C₁₈H₃₄O₂ 6% and linoleic C₁₈H₃₂O₂ 2%.

a

Find the percentage of omega-3 fatty acids. (1 mark)

b

Why is coconut oil resistant to oxidative rancidity? (1 mark)

c

Coconut oil reacts in the body to give fatty acids and one other product. Name it. (1 mark)

d

Give the molecular formula of the molecule in coconut oil that gives three caprylic acid molecules. (1 mark)

e

Explain why vitamin E (a large, mostly hydrocarbon molecule with one OH) dissolves in coconut oil. (2 marks)

f

A coconut oil fuel cell uses KOH(aq). i. Write the cathode half-equation. (1 mark)

ii

State one design feature of the cathode that helps efficiency. (1 mark)

g

Why might a fuel cell be better than burning coconut oil? (1 mark)

h

Give one reason, not environmental, why coconut oil may be impractical in a fuel cell. (2 marks)

Show worked solution

a. [1 mark]

VCAA judged that the question contained conflicting information, so every student received this mark. On the Data Book's definitions, none of the listed acids is omega-3 (linoleic acid, C₁₈H₃₂O₂, is omega-6), giving 0%.

b. [1 mark]

It consists mainly of saturated fatty acids, with very few C=C double bonds, which are the sites where oxidation occurs.

c. [1 mark]

Glycerol (the fats are hydrolysed).

d. [1 mark]

Three caprylic acids condense with glycerol, losing three water molecules:

$$3\text{C}_8\text{H}_{16}\text{O}_2 + \text{C}_3\text{H}_8\text{O}_3 - 3\text{H}_2\text{O} = \text{C}_{27}\text{H}_{50}\text{O}_6.$$

e. [2 marks]

Vitamin E is mostly non-polar (a large hydrocarbon structure with only one OH), and coconut oil is non-polar. Dispersion forces can form between the two, so vitamin E dissolves ("like dissolves like").

f. i. [1 mark]

$\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq)$

ii. [1 mark]

Any one: a porous electrode (large surface area), or a catalyst coating on the electrode.

g. [1 mark]

A fuel cell converts chemical energy to electrical energy more efficiently, with less energy lost as heat, than combustion.

h. [2 marks]

A reason plus its practical effect, for example: coconut oil is viscous (or solid at cool temperatures), so it would not flow well through the cell; or it is non-polar and insoluble in water, so it cannot mix with the aqueous KOH electrolyte to react effectively.

From the report. Part d (about 20% correct) was hard: students struggled to reassemble the triglyceride. In f.i, states from the Data Book and a single-headed arrow were required. In h, many wrongly believed fuel cells need gaseous fuels.

Question 2 (12 marks)

Lactose is hydrolysed by the enzyme lactase into two monosaccharides.

a

Name the circled linkage in lactose. (1 mark)

b. i

Explain what the GI of a food indicates. (2 marks)

ii

State why the breakdown of lactose contributes to its GI. (1 mark)

c

Proline is in cow's milk and in lactase. i. Why is proline not an essential amino acid? (1 mark)

ii

Draw proline in acidic solution. (1 mark)

iii

Identify two types of enzyme-catalysed reactions involving proline in the body. (2 marks)

iv

Explain why proline has two optical isomers. (2 marks)

v

Explain why lactase must contain the correct optical isomer of proline to work. (2 marks)

Show worked solution

a. [1 mark]

Glycosidic link.

b. i. [2 marks]

The glycaemic index measures how quickly a food raises blood glucose levels after it is eaten, relative to glucose (which has a GI of 100), typically over about 2 hours.

ii. [1 mark]

Hydrolysis of lactose produces glucose (and galactose), and the glucose enters the bloodstream.

c. i. [1 mark]

The body can synthesise proline, so it does not have to come from the diet.

ii. [1 mark]

In acid, the ring nitrogen is protonated: it carries two hydrogens and a positive charge (-NH₂⁺- in the ring), and the carboxyl group stays as -COOH.

iii. [2 marks]

Any two: condensation reactions (proline joining other amino acids to build proteins such as enzymes); hydrolysis reactions (releasing proline when proteins are digested); oxidation reactions (breaking proline down).

iv. [2 marks]

Proline has a chiral carbon (the α-carbon), bonded to four different groups: H, COOH, the ring nitrogen and a ring CH₂. It can therefore exist as two non-superimposable mirror images (enantiomers).

v. [2 marks]

Proline is part of the enzyme's primary structure. The wrong optical isomer would change how the chain folds, and therefore the shape of lactase's active site. Lactose would then no longer fit the active site (its shape would not be complementary), so lactose in milk would not be digested effectively.

From the report. In b.ii, many wrongly thought lactose hydrolyses to two glucose units (it gives glucose and galactose). In c.ii the positive charge had to be on the nitrogen, not on the hydrogens. In c.v many treated proline as the substrate instead of part of the enzyme.

Question 3 (12 marks)

a

Propane is used as a barbecue fuel. i. Calculate the energy released when 140.0 g of propane burns completely. (1 mark)

ii

Write the thermochemical equation for its complete combustion at SLC. (2 marks)

iii

Sketch the energy profile. (1 mark)

iv

How would the profile for incomplete combustion differ? Justify. (2 marks)

b

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). i. 96.0 g of ethanol burns and the CO₂ is captured. Find the tank volume needed at 15.0 °C and 100.0 kPa. (3 marks)

ii

Is the flashpoint of ethanol ($M = 46.0$) higher or lower than that of propane ($M = 44.0$)? Explain with reference to their structures. (3 marks)

Show worked solution

a. i. [1 mark]. From the Data Book, propane's heat of combustion is 50.5 kJ g⁻¹:

$$E = 50.5 \times 140.0 = 7.07 \times 10^3 \text{ kJ}.$$

ii. [2 marks].

$$\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \qquad \Delta H = -2220 \text{ kJ mol}^{-1}$$

iii. [1 mark]. An exothermic profile: products well below reactants, with an activation energy hump.

iv. [2 marks]. The products (such as CO or C, with H₂O) would sit higher than CO₂ and H₂O, though still below the reactants, so $\Delta H$ is smaller in size. Carbon is only partially oxidised, so less energy is released.

b. i. [3 marks].

$$n(\text{C}_2\text{H}_5\text{OH}) = \frac{96.0}{46.0} = 2.087 \text{ mol}, \qquad n(\text{CO}_2) = 2 \times 2.087 = 4.17 \text{ mol}$$

$$V = \frac{nRT}{P} = \frac{4.17 \times 8.31 \times 288}{100.0} \approx 99.9 \text{ L}.$$

ii. [3 marks]. Ethanol's flashpoint is higher. Although the molar masses are almost equal, ethanol has a hydroxy (-OH) group, so its molecules are held by hydrogen bonds as well as dispersion forces, whereas propane has only weak dispersion forces. More energy (a higher temperature) is needed to vaporise enough ethanol to form a flammable mixture with air.

From the report. In a.i, kJ g⁻¹ is the wrong unit for an amount of energy. In a.ii, $-50.5$ kJ g⁻¹ does not belong in a thermochemical equation, and at SLC water must be a liquid. In a.iv, "because it is incomplete combustion" repeats the question; the reason is incomplete oxidation.

Question 4 (7 marks)

But-1-ene reacts with steam (H₃PO₄ catalyst, 300 °C) to give two products, Q and R. Only R can be oxidised by acidified dichromate to butanoic acid; incomplete oxidation of R gives T. Butanoic acid reacts with ethanamine to give S.

a. i

Name the type of reaction that forms Q and R. (1 mark)

ii

Name Q. (1 mark)

b

Draw the structural formula of T. (1 mark)

c

Draw the skeletal formula of S. (2 marks)

d

Butanoic acid and ethanol form an ester with $M = 116.0$ g mol⁻¹. Calculate the atom economy. (2 marks)

Show worked solution

a. i. [1 mark]

Addition (hydration).

ii. [1 mark]

Butan-2-ol. Water can add across the double bond either way; R is butan-1-ol (a primary alcohol, which oxidises to the acid), and Q is the secondary alcohol butan-2-ol.

b. [1 mark]

T is the aldehyde butanal, CH₃CH₂CH₂CHO: draw all atoms and bonds, with the C=O at the end carbon bonded to one H.

c. [2 marks]

S is the amide N-ethylbutanamide, CH₃CH₂CH₂CONHCH₂CH₃:

d. [2 marks]. The ester is ethyl butanoate. Butanoic acid (88.0) + ethanol (46.0) → ester (116.0) + water:

$$\text{atom economy} = \frac{116.0}{88.0 + 46.0} \times 100\% = \frac{116.0}{134.0} \times 100\% \approx 86.6\%.$$

From the report. In c, one mark was for a correct amide link and one for a correct skeletal diagram of the whole molecule. In d, 87% or 86.57% was also fine; this was not a significant-figure question.

Question 5 (7 marks)

CH₄(g) + 2H₂O(g) ⇌ CO₂(g) + 4H₂(g), $\Delta H = +165$ kJ mol⁻¹, at 300 °C with a catalyst.

a

State a source of CH₄. (1 mark)

b

25.0 mol CH₄ and 25.0 mol H₂O are sealed in an empty 100.0 L container at 300 °C; at equilibrium there is 6.12 mol H₂. i. Write the equilibrium expression. (1 mark)

ii

Calculate $K$. (4 marks)

c

How does lowering the temperature change the shape of the Maxwell-Boltzmann distribution? (1 mark)

Show worked solution

a. [1 mark]. Any one: natural gas, coal seam gas, biogas (anaerobic digestion of organic waste), or cattle.

b. i. [1 mark].

$$K = \frac{[\text{CO}_2][\text{H}_2]^4}{[\text{CH}_4][\text{H}_2\text{O}]^2}$$

ii. [4 marks]. 6.12 mol of H₂ formed, so $x = \tfrac{6.12}{4} = 1.53$ mol reacted:

	CH₄	H₂O	CO₂	H₂
Initial (mol)	25.0	25.0	0	0
Change (mol)	$-1.53$	$-3.06$	$+1.53$	$+6.12$
Equilibrium (mol)	23.47	21.94	1.53	6.12
Equilibrium (M)	0.2347	0.2194	0.0153	0.0612

$$K = \frac{0.0153 \times 0.0612^4}{0.2347 \times 0.2194^2} \approx 1.90 \times 10^{-5} \text{ M}^2.$$

c. [1 mark]. The curve becomes narrower and taller, with its peak shifted to lower energy (the area under it is unchanged).

From the report. Part b.ii was generally well done with clear, structured working, and the units of $K$ earned a mark. In c, "narrower" and "taller" both had to be mentioned.

Question 6 (7 marks)

An experimental secondary (flow) cell discharges by Fe₂(SO₄)₃(aq) + C₁₄H₁₀O₈S₂(aq) → 2FeSO₄(aq) + C₁₄H₈O₈S₂(aq) + HSO₄⁻(aq) + H⁺(aq), with carbon electrodes and a membrane.

a

State the energy transformation during discharge. (1 mark)

b

Which reactant is the oxidising agent during discharge? Justify with oxidation numbers. (2 marks)

c. i

Write the half-equation in the C₁₄H₈O₈S₂/C₁₄H₁₀O₈S₂ half-cell during recharge. (1 mark)

ii

State the polarity of that electrode during recharge. (1 mark)

iii

Explain how electrode polarity is established during recharge. (2 marks)

Show worked solution

a. [1 mark]

Chemical energy to electrical energy.

b. [2 marks]

Fe₂(SO₄)₃ (the Fe³⁺ ions): iron's oxidation number falls from +3 in Fe₂(SO₄)₃ to +2 in FeSO₄, so Fe³⁺ is reduced and is the oxidising agent.

c. i. [1 mark]

On discharge the organic compound is oxidised (loses 2H), so recharge reverses this:

$$\text{C}_{14}\text{H}_8\text{O}_8\text{S}_2(aq) + 2\text{H}^+(aq) + 2e^- \rightarrow \text{C}_{14}\text{H}_{10}\text{O}_8\text{S}_2(aq)$$

ii. [1 mark]. Negative.

iii. [2 marks]. The polarity of each electrode does not change between discharge and recharge: it is fixed by the relative strengths of the oxidant and reductant in the cell (the organic half-cell is the negative anode on discharge). For recharge, the external power supply's negative terminal is connected to the negative electrode, forcing electrons in; this reverses the process at each electrode (reduction now happens at the negative electrode and oxidation at the positive one), regenerating the original reactants.

From the report. In c.iii many wrongly said the polarity of the electrodes "swaps" during recharge; the polarity stays the same and the processes reverse.

Question 7 (10 marks)

Molecule V contains only C, H and one O. Its mass spectrum has the molecular ion at m/z = 86 with a small peak at 87. Its ¹H NMR spectrum has three signals, including a doublet at 1.1 ppm; its ¹³C NMR spectrum has four peaks.

a. i

State the molecular formula of V, with justification from the mass spectrum. (2 marks)

ii

Why is there a small peak at m/z = 87? (1 mark)

b

What does the doublet at 1.1 ppm tell you? (1 mark)

c

Draw a structural formula of V consistent with parts a to c. (3 marks)

d

Explain why one molecule absorbs different frequencies of infrared radiation. (3 marks)

Show worked solution

a. i. [2 marks]

The molecular ion (parent) peak is at m/z = 86, so $M = 86$ g mol⁻¹. With one O (16), the hydrocarbon part is 70, which fits C₅H₁₀: the formula is C₅H₁₀O.

ii. [1 mark]

Some molecules contain a ¹³C isotope, giving an ion one mass unit heavier.

b. [1 mark]

By the $n + 1$ rule, the hydrogens giving this doublet are on a carbon next to a carbon with one hydrogen.

c. [3 marks]

3-methylbutan-2-one, CH₃COCH(CH₃)₂. It has three hydrogen environments (CH₃CO, the CH and two equivalent CH₃ groups, which give the 6H doublet) and four carbon environments (the C=O, the CH₃ next to it, the CH and the two equivalent CH₃). In skeletal form (the exam asked for a full structural diagram showing every atom):

d. [3 marks]. Each type of bond (C-H, C=O, C-C and so on) absorbs infrared radiation at its own frequency, because the frequency at which a bond vibrates depends on its strength and the masses of the atoms it joins. Different kinds of vibration (stretching and bending) also absorb at different frequencies, so one molecule gives many absorptions.

From the report. The third mark in c was only for a correct full structural diagram of 3-methylbutanone. In d, the first mark needed the link between different bonds or functional groups and different absorbed frequencies.

Question 8 (16 marks)

A student burns crushed dry beans under an aluminium can holding 300.0 mL of water, measuring the temperature every 30 seconds, inside a larger can with air holes on a heatproof mat, and repeats this with soybeans.

a

Why were the beans crushed? (2 marks)

b

Identify the dependent variable. (1 mark)

c

Identify a step that manages a safety hazard and explain how. (2 marks)

d

1.11 g of broad-bean powder heats the water from 25.0 °C to 32.1 °C. Calculate the energy content in kJ g⁻¹. (4 marks)

e

Dry soybeans contain 36 g protein, 20 g fat and 29 g carbohydrate per 100 g. i. Calculate their expected heat of combustion in kJ g⁻¹. (2 marks)

ii

The experiment gave 35 kJ g⁻¹. Suggest an error that could explain the difference and how. (2 marks)

f

Suggest one improvement for more accurate results and explain it. (2 marks)

g

Write a valid conclusion. (1 mark)

Show worked solution

a. [2 marks]

Crushing increases the surface area, so the beans burn faster and more completely.

b. [1 mark]

The temperature change of the water (or the calculated energy content). Just "temperature" is not enough.

c. [2 marks]

For example: step 4 uses a heatproof mat, which prevents the burning beans from setting the bench on fire; or step 8 waits until the water is below 30 °C, reducing the risk of scalds.

d. [4 marks]

$\Delta T = 32.1 - 25.0 = 7.1$ °C. Using the density of water from the 2023 Data Book (0.997 g mL⁻¹), $m = 300.0 \times 0.997 = 299.1$ g.

$$q = mc\Delta T = 299.1 \times 4.18 \times 7.1 = 8877 \text{ J} = 8.88 \text{ kJ}$$

$$\text{energy content} = \frac{8.88}{1.11} \approx 8.0 \text{ kJ g}^{-1}.$$

e. i. [2 marks].

$$36(17) + 20(37) + 29(16) = 1816 \text{ kJ per } 100 \text{ g} \approx 18 \text{ kJ g}^{-1}.$$

ii. [2 marks]

The experimental value is too high, so the error must make $\Delta T$ too large or the mass too small. For example, the thermometer rested on the bottom of the can, which was hotter than the water, so the recorded temperature change (and energy content) was too high; or the water was not stirred, so a local hot spot gave too large a $\Delta T$.

f. [2 marks]

For example: insulate the outer can (or shield from draughts) so that more of the energy released reaches the water, making $\Delta T$ and the energy content closer to the true value; or repeat the trials and discard outliers. (Sealing the can or closing the air holes is wrong: it would cause incomplete combustion.)

g. [1 mark]

Based on the results, soybeans have a higher energy content than broad beans (or: no valid conclusion can be drawn because of the error in the soybean trial).

From the report. In d, fewer than 5% of students earned all four marks; the answer needed the mass of water and two significant figures. In e.i, many gave 1816 kJ g⁻¹, missing that the data were per 100 g. In e.ii, the error had to be linked to a $\Delta T$ that was too high or a mass that was too low.

Question 9 (8 marks)

a. Copper that is 99% pure (containing Zn, Pb and Au) is refined by electrolysis, with the impure copper as the anode, pure copper as the cathode and 1 M CuSO₄ with 1 M H₂SO₄ as the electrolyte. Discuss how this purifies the copper, defining electrolysis and explaining what happens to the Au, Zn and Pb. (5 marks)

b. Explain how the Faraday constant and Faraday's first law can find the amount of Br₂ produced by electrolysing molten KBr. (3 marks)

Show worked solution

a. [5 marks]. A strong answer covers:

• Definition. Electrolysis uses an electric current (an external power supply) to drive a non-spontaneous redox reaction.
• At the anode (impure copper). Zn and Pb are stronger reductants than Cu (lower on the electrochemical series), so they are oxidised along with the copper: Zn → Zn²⁺ + 2e⁻, Pb → Pb²⁺ + 2e⁻ and Cu → Cu²⁺ + 2e⁻.
• Gold. Au is a much weaker reductant than Cu, so it is not oxidised at the applied voltage; it falls from the anode and collects as anode "mud".
• At the cathode (pure copper). Cu²⁺ is a stronger oxidant than Zn²⁺ and Pb²⁺, so only Cu²⁺ is reduced (Cu²⁺ + 2e⁻ → Cu), depositing pure copper; Zn²⁺ and Pb²⁺ stay in solution.
• Voltage control. The applied voltage must be high enough to oxidise copper but too low to oxidise gold (or to reduce Zn²⁺ and Pb²⁺).

b. [3 marks]. Measure the current $I$ and time $t$ to find the charge, $Q = It$. Dividing by the Faraday constant gives the amount of electrons, $n(e^-) = \dfrac{Q}{F}$. From 2Br⁻ → Br₂ + 2e⁻, each mole of Br₂ needs 2 mol of electrons, so $n(\text{Br}_2) = \tfrac12 n(e^-)$.

From the report. Part a was demanding: very few students earned 4 or 5 marks. Each mark required a specific idea: the definition, Zn and Pb oxidised at the anode, Au not oxidised, only Cu²⁺ reduced at the cathode, and the voltage window.

General advice from the 2023 report

• Annotate application questions: write out equations or draw structures before choosing an answer in Section A.
• Show working in calculations. Most students did, which let markers award consequential marks.
• Use the Data Book's exact states and values in thermochemical and half-equations, with single-headed arrows for reactions that occur.
• Read the stem closely, for example "per 100 g" in Question 8e and "at SLC" in Question 3a.